Question Number 32768 by Rasheed.Sindhi last updated on 01/Apr/18
$$\mathrm{Prove}\:\mathrm{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ca} \\ $$$$\forall\:{a},{b},{c}\in\mathbb{R} \\ $$
Answered by mrW2 last updated on 02/Apr/18
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\left({a}−{b}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} \geqslant\mathrm{2}{ab} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant\mathrm{2}{bc} \\ $$$${c}^{\mathrm{2}} +{a}^{\mathrm{2}} \geqslant\mathrm{2}{ca} \\ $$$$\Sigma: \\ $$$$\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \geqslant{ab}+{bc}+{ca} \\ $$
Commented by Joel578 last updated on 02/Apr/18
$$\mathrm{In}\:\mathrm{3rd}\:\mathrm{line},\:\mathrm{did}\:\mathrm{you}\:\mathrm{mean}\: \\ $$$${c}^{\mathrm{2}} \:+\:{a}^{\mathrm{2}} \:\geqslant\:\mathrm{2}{ca}\:\:? \\ $$
Commented by Rasheed.Sindhi last updated on 02/Apr/18
$$\mathrm{Th}\forall\mathrm{n}\Bbbk\mathrm{s}\:\mathrm{Sir}! \\ $$
Commented by mrW2 last updated on 02/Apr/18
$${yes}.\:{thanks}! \\ $$